Problem: Which of the following numbers is a multiple of 2? ${41,52,61,99,103}$
Solution: The multiples of $2$ are $2$ $4$ $6$ $8$ ..... In general, any number that leaves no remainder when divided by $2$ is considered a multiple of $2$ We can start by dividing each of our answer choices by $2$ $41 \div 2 = 20\text{ R }1$ $52 \div 2 = 26$ $61 \div 2 = 30\text{ R }1$ $99 \div 2 = 49\text{ R }1$ $103 \div 2 = 51\text{ R }1$ The only answer choice that leaves no remainder after the division is $52$ $ 26$ $2$ $52$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $2$ are contained within the prime factors of $52$ $52 = 2\times2\times13 2 = 2$ Therefore the only multiple of $2$ out of our choices is $52$. We can say that $52$ is divisible by $2$.